Question

what must be added to the number 6,10,14 and 22 so that they are are in proportion​

Answer 1

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$let \: the \: required \: number \: be \: x \: then \\ 6 + x \: 10 + x \: 14 + x \: 22 + x \: are \: in \: proportion$

$⟹product \: of \: extremes = product \: of \: means$

$⟹(6 + x)(22 + x) = (10 + x)(14 + x)$

$⟹132 + 6x + 22x + {x}^{2} = 140 + 10x + 14x + {x}^{2}$

$⟹132 + 28x = 140 + 24x$

$⟹28x - 24x = 140 - 132$

$⟹4x = 8$

$⟹x = \frac{8}{4} = 2.$

Hence,the required number is 2.

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Answer 2

$\huge\tt{\underline{Solution}}$

let there quired number be x then

6+x10+x14+x22+x are in proportion

⟹product of extremes = product of means

$\begin{gathered}(6 + x)(22 + x) = (10 + x)(14 + x) \\\end{gathered}$

$\begin{gathered}132 + 6x + 22x + {x}^{2} = 140 + 10x + 14x + {x}^{2} \\\end{gathered}$

$\sf{132 + 28x = 140 + 24x}$

$\sf{28x - 24x = 140 - 132}$

$\sf{4x = 8}$

$\sf{x = \frac{8}{4}} = 2$

Hence,the required number is 2.